Question

Among recent college graduates who had at least one student loan during college, we took a Simple Random Sample of size 18. The average monthly student loan payment for those in the sample was $363.07; the (sample) standard deviation was $49.05.

Use the sample data to find a 99% confidence interval for the average monthly student loan payment among all recent college graduates who had at least one student loan during college.Round all answers to the nearest cent (two decimal places).The margin of error for this confidence interval is: $   

The 99% confidence interval is:

$ < μ < $

Answer 1

sample mean, xbar = 363.07
sample standard deviation, s = 49.05
sample size, n = 18
degrees of freedom, df = n - 1 = 17

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.898

Margin of error,
ME = tc * s/sqrt(n)
ME = 2.898 * 49.05/sqrt(18)
ME = 33.50

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (363.07 - 2.898 * 49.05/sqrt(18) , 363.07 + 2.898 * 49.05/sqrt(18))
CI = (329.57 , 396.57)

$329.57 < mu < $396.57